B = \(\frac{3^2}{2.4}+\frac{3^2}{4.6}+\frac{3^2}{6.8}+...+\frac{3^2}{198.200}\)

B = \(\frac{3^2}{2}.\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{3^2}{2}.\left(\frac{1}{4}-\frac{1}{6}\right)+\frac{3^2}{2}.\left(\frac{1}{6}-\frac{1}{8}\right)+...+\frac{3^2}{2}.\left(\frac{1}{198}-\frac{1}{200}\right)\)

B = \(\frac{3^2}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{198}-\frac{1}{200}\right)\)

B = \(\frac{9}{2}.\left(\frac{1}{2}-\frac{1}{200}\right)\)

B = \(\frac{9}{2}.\frac{99}{200}\)

B = \(\frac{891}{400}\)

D = 1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + ... + 48 x 49

3D = 1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3 + 4 x 5 x 3 + ... + 48 x 49 x 3

3D = 1 x 2 x 3 + 2 x 3 x ( 4 - 1 ) + 3 x 4 x ( 5 - 2 ) + 4 x 5 x ( 6 - 3 ) + ... + 48 x 49 x ( 50 - 47 )

3D = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + 4 x 5 x 6 - 3 x 4 x 5 + ... + 48 x 49 x 50 - 47 x 48 x 49

3D = 48 x 49 x 50

D = ( 48 x 49 x 50 ) : 3

D = 39200

E = 1² + 2² + 3² + ... + 48²

E = 1 x 1 + 2 x 2 + 3 x 3 + ... + 48 x 48

E = 1 x ( 2 - 1 ) + 2 x ( 3 - 1 ) + 3 x ( 4 - 1 ) + ... + 48 x ( 49 - 1 )

E = 1 x 2 - 1 + 2 x 3 - 2 + 3 x 4 - 3 + ... + 48 x 49 - 49

E = ( 1 x 2 + 2 x 3 + 3 x 4 + ... + 48 x 49 ) - ( 1 + 2 + 3 + ... + 49 )

Ta tính được vế trong ngoặc thứ nhất là 39200 , còn vế trong ngoặc thứ hai là 1225

thay vào ta được :

E = 39200 - 1225

E = 37975 

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

\(\Rightarrow A=1-\frac{1}{2^{100}}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{1}{3}-\frac{1}{15}+2\left(1-\frac{1}{10}\right)\)

\(=\frac{4}{15}+\frac{9}{5}\)

\(=\frac{31}{15}\)

              Bài làm :

Ta có :

\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{13\times15}+\frac{2}{1\times2}+\frac{2}{2\times3}+...+\frac{2}{9\times10}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{1}{3}-\frac{1}{15}+2\left(1-\frac{1}{10}\right)\)

\(=\frac{31}{15}\)

\(\frac{B}{2}=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{100\cdot101}\)

\(\frac{B}{2}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}\)

\(\frac{B}{2}=\frac{100}{101}\)

\(B=\frac{200}{101}\)

B = \(2\left(\frac{1}{1x2}+\frac{1}{2x3}+....+\frac{1}{100x101}\right)\)

B = \(2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}...+\frac{1}{101}\right)\)

B = \(2\left(1-\frac{1}{101}\right)\)

B = \(2x\frac{100}{101}\)

B = \(\frac{200}{101}\)

\(B=\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+...+\frac{2}{100\cdot101}\)

\(B=2\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{100\cdot101}\right)\)

\(B=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)

\(B=2\left(1-\frac{1}{101}\right)\)

\(B=2\cdot\frac{100}{101}\)

\(B=\frac{200}{101}\)

\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{19.20}\)

\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{19.20}\right)\)

\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{19}-\frac{1}{20}\right)\)

\(=2.\left(1-\frac{1}{20}\right)\)

\(=2.\frac{19}{20}=\frac{19}{10}\)

\(17.8+51.4=34.4+51.4=4\left(51+34\right)=4.84=336\) \(2.2.3.5.19=\left(2.5\right).\left(3.19\right).2=10.2.57=570.2=1140\) \(54.275+825.15+275=54.275+45.275+275=275\left(54+45+1\right)=100.275=27500\) \(\frac{167.198+98}{198.168-100}=\frac{167.198+98}{198.167+198-100}=\frac{167.198+98}{167.198+98}=1\)

\(\frac{1}{n}-\frac{1}{n+k}=\frac{k}{n\left(n+k\right)}\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{2020}=1-\frac{1}{2020}=\frac{2019}{2020}\)

a) 17 x 8 + 51 x 4

= 17 x 4 x 2 + 17 x 3 x 4

= 17 x 4 x ( 2 + 3 )

= 14 x 4 x 5

= 14 x 20

= 280

b) 2 x 2 x 3 x 5 x 19

= ( 2 x 5 ) x ( 3 x 19 ) x 2

= 10 x 57 x 2

= 570 x 2

= 1140

c) 54 x 275 + 825 x 15 + 275

= 54 x 275 + 275 x 3 x 15 + 275 x 1

= 54 x 275 + 275 x 45 + 275 x 1

= 275 x ( 54 + 45 + 1 )

= 275 x 100

= 27500

d) 100 - 99 + 98 - 97 + 96 - 95 + 94 - 93 + ... + 4 - 3 + 2

= (100 - 99) + (98 - 97) + (96 - 95) + (94 - 93) + ... + (4 - 3) + 2

= (1 + 1 + ... + 1) + 2

( 49 số 1 )

= 49 + 2

= 51

k) 1,5 + 2,5 + 3,5 + 4,5 + 5,5 + 6,5 + 7,5 + 8,5

= ( 1,5 + 8,5 ) + ( 2,5 + 7,5 ) + ( 3,5 + 6,5 ) + ( 4,5 + 5,5 )

= 10 + 10 + 10 + 10

= 40

m) Đề bn xem ở trên

= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{1}-\frac{1}{10}\)

\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)

a) \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{5}-\frac{1}{10}\)

\(=\frac{1}{10}\)

b) \(\frac{2}{10.12}+\frac{2}{12.14}+\frac{2}{14.16}+...+\frac{2}{998.1000}\)

\(=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}-\frac{1}{16}+...+\frac{1}{998}-\frac{1}{1000}\)

\(=\frac{1}{10}-\frac{1}{1000}\)

\(=\frac{99}{1000}\)

c) \(\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{69.90}\)

\(=4.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{89.90}\right)\)

\(=4.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{89}-\frac{1}{90}\right)\)

\(=4.\left(1-\frac{1}{90}\right)\)

\(=4.\frac{89}{90}\)

\(=\frac{178}{45}\)

_Chúc bạn học tốt_

a, \(=\frac{1}{10}\)

a) 1/10

b) ............

c)............

Mình giúp rồi t.i.c.k mình :v